Find the real solution(s) of the given equation. \( 4 x^{4}+25=29 x^{2} \)

To solve the equation \( 4x^{4} + 25 = 29x^{2} \), we first rearrange it to set it to zero: \[ 4x^{4} - 29x^{2} + 25 = 0 \] Next, we can introduce a substitution to simplify our equation. Let \( y = x^{2} \). This transforms our equation into a quadratic form: \[ 4y^{2} - 29y + 25 = 0 \] Now, we can apply the quadratic formula: \[ y = \frac{-b \pm \sqrt{b^{2} - 4ac}}{2a} \] Here, \( a = 4 \), \( b = -29 \), and \( c = 25 \): \[ y = \frac{29 \pm \sqrt{(-29)^{2} - 4 \cdot 4 \cdot 25}}{2 \cdot 4} \] \[ y = \frac{29 \pm \sqrt{841 - 400}}{8} \] \[ y = \frac{29 \pm \sqrt{441}}{8} \] \[ y = \frac{29 \pm 21}{8} \] Now we calculate the two potential values for \( y \): 1. \( y = \frac{50}{8} = 6.25 \) 2. \( y = \frac{8}{8} = 1 \) Recalling our substitution \( y = x^{2} \), we will solve for \( x \): 1. For \( y = 6.25 \): \[ x^{2} = 6.25 \Rightarrow x = \pm \sqrt{6.25} = \pm 2.5 \] 2. For \( y = 1 \): \[ x^{2} = 1 \Rightarrow x = \pm \sqrt{1} = \pm 1 \] Gathering all the real solutions, we have: \[ x = 2.5, -2.5, 1, -1 \] So the real solutions of the given equation are \( x = 2.5, -2.5, 1, -1 \).